QJEEM21Ma4e1d: JEE Main 2021 Mathematics - Class 11 JEEM21Ma4e1d Solution: Let and be the roots of x 2 6x 2 = 0. If a n = n n for n 1,…

Question

Verified Solution • 2025–26 Session moderate

Let

\alpha

and

\beta

be the roots of x²

-

6x

-

2 = 0. If a_n =

\alpha

ⁿ

-

\beta

ⁿ for n

\ge

1, then the value of

{{{a_{10}} - 2{a_8}} \over {3{a_9}}}

is :

Accepted Answer

Given,

\alpha

and

\beta

be the roots of

{x^2} - 6x - 2 = 0

\begin{matrix} α + β = 6 \\ α β = - 2 \end{matrix}

and

{\alpha ^2} - 6\alpha - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha

{\beta ^2} - 6\beta - 2 = 0 \Rightarrow {\beta ^2} - 2 = 6\beta

{{{a_{10}} - 2{a_8}} \over {3{a_9}}} = {{\left( {{\alpha ^{10}} - {\beta ^{10}}} \right) - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}

= {{\left( {{\alpha ^{10}} - 2{\alpha ^8}} \right) - \left( {{\beta ^{10}} - 2{\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}

Now,

= {{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}

= {{{\alpha ^8}(6\alpha ) - {\beta ^8}(6\beta )} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {{6\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {6 \over 3} = 2

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